Practice Ci with instalments - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   Rs. 16,820 is divided between two brothers of age 27 years and 25 years. They invested their money at 5% per annum compound interest in such a way that both will receive equal money at the age of 40 years. The share (in Rs.) of elder brother is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Share of elder brother = Rs.x (let)

Share of younger brother = Rs.(16820 - x)

A = P$(1 + R/100)^T$

According to the question,

$x(1 + 5/100)^{13} = (16820 - x)(1 + 5/100)^{15}$

$x = (16820 - x)(1 + 1/20)^2$

$x = (16820 - x)(21/20)^2$

$(20/21)^2x = 16820 - x$

${400x}/441 + x$ = 16820

${400x + 441x}/441$ = 16820

841x = 16820 × 441

$x = {16820 × 441}/841$ = Rs.8820

Q-2)   A sum of Rs.210 was taken as a loan. This is to be paid back in two equal instalments. If the rate of interest be 10% compounded annually, then the value of each instalment is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 9(i),

Let the value of each instalment be Rs.x

Principal = Present worth of Rs.x due 1 year hence, present worth of Rs. x due 2 years hence

210 = $x/(1 + R/100) + x/(1 + R/100)^2$

210 = $x/(1 + 10/100) + x/(1 + 10/100)^2$

210 = $x/{1 + 1/10} + x/(1 + 1/10)^2$

210 = $x/{11/10} + x/(11/10)^2$

210 = ${10x}/11 + {100x}/121$

210 = ${110x + 100x}/121$

210 × 121 = 210 x

$x = {210 × 121}/210$ = Rs.121

Q-3)   A builder borrows Rs.2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be ?

(a)

(b)

(c)

(d)

Explanation:

A = Rs.2550

R = 4% per annum

n = 2 years

Let each of the two equal instalments be x

Present worth = $\text"Instalment"/(1 + r/100)^n$

$P_1 = x/(1 + 4/100)^1 = x/{1 + 1/25} = x/{26/25}$

or $P_1 = 25/26x$

Similarly,

$P_2 = (25/26)^2x = 625/676x$

$P_1 + P_2$ = A

$25/26x + 625/676x$ = 2550

${(650 + 625)x}/676 = 2550$

$1275/676x = 2550$

x = 2550 $× 676/1275$ ⇒ x = Rs.1352

Using Rule 9,
If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then
(i) For n = 2, Each annual installment
= $p/{(100/{100 + r}) + (100/{100 + r})^2$
(ii) For n = 3, Each annual installment
= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$

Here, P = Rs.2550, n = 2, r = 4%

Each instalment

= $p/{(100/{100 + r}) + (100/{100 + r})^2$

= $2550/{(100/{100 + 4}) + (100/{100 + 4})^2$

= $2550/{100/104 + (100/104)^2}$

= $2550/{100/104(1 + 100/104)}$

= $2550/{100/104 (204/104)}$

= ${2550 × 104 × 104}/20400 $= Rs.1352

Q-4)   Kamal took Rs.6800 as a loan which along with interest is to be repaid in two equal annual instalments. If the rate of interest is 12$1/2$%, compounded annually, then the value of each instalment is

(a)

(b)

(c)

(d)

Explanation:

Let the annual instalment be x

A = P$(1 + R/T)^T$

$x = P_1(1 + 25/200)$

$x = P_1 × 9/8$

$P_1 = 8/9x$

Similarly, $P_2 = 64/81x$

$P_1 + P_2$ = 6800

$8/9x + 64/81x$ = 6800

${72x + 64x}/81 = 6800$

${136x}/81 = 6800$

$x = {6800 × 81}/136$ = Rs.4050

Using Rule 9(i),

Here, P = Rs.6800, R = $25/2$%, n = 2

Each instalment

= $p/{(100/{100 + r}) + (100/{100 + r})^2$

= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$

= $6800/{200/225 + (200/225)^2}$

= $6800/{200/225(1 + {200/225})}$

= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050

Q-5)   A loan of Rs.12,300 at 5% per annum compound interest, is to be repaid in two equal annual instalments at the end of every year. Find the amount of each instalment.

(a)

(b)

(c)

(d)

Explanation:

Using Rule 9(i),

Let each instalment be x.

$x/(1 + 5/100) + x/(1 + 5/100)^2 = 12300$

${20x}/21 + (20/21)^2x = 12300$

${20x}/21(1 + 20/21)$ = 12300

${20x}/21 × 41/21 × x = 12300$

$x = {12300 × 21 × 21}/{20 × 41}$ ⇒ x = 6615

Q-6)   A man buys a scooter on making a cash down payment of Rs.16224 and promises to pay two more yearly instalments of equivalent amount in next two years. If the rate of interest is 4% per annum, compounded yearly, the cash value of the scooter, is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,
If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,
A=P$(1 + r/100)^n$, C.I. = A - P
C.I. = P$[(1 + r/100)^n - 1]$

Let principal (present worth) for first year be $P_1$ and that for two years be $P_2$.

16224 = $P_1(1 + 4/100)$

16224 = $P_1(1 + 1/25) = {26P_1}/25$

$P_1 = {16224 × 25}/26$ = Rs.15600

Again,

$16224 = P_2(1 + 4/100)^2$

16224 = $P_2(26/25)^2 = {676P_2}/625$

$P_2 = {16224 × 625}/676$ = 15000

Cash value of the scooter

= Rs.(16224 + 15600 + 15000) = Rs.46824

Q-7)   A sum of money is paid back in two annual instalments of Rs. 17, 640 each, allowing 5% compound interest compounded annually. The sum borrowed was

(a)

(b)

(c)

(d)

Explanation:

Using Rule 9(i),

Sum borrowed = Present worth of Rs.17640 due 1 year hence + Present worth of Rs.17640 due 2 years hence

= Rs.$(17640/{(1 + 5/100)} + 17640/{(1 + 5/100)^2})$

= Rs.$(17640 × 20/21 + 17640 × 20/21 × 20/21)$

= Rs.(16800 + 16000) = Rs.32800

Q-8)   Mr. Dutta desired to deposit his retirement benefit of Rs. 3 lakhs partly to a post office and partly to a bank at 10% and 6% interests respectively. If his monthly interest income was Rs. 2000, then the difference of his deposits in the post office and in the bank was :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Let the amount deposited in Post Office be Rs.x lakhs.

Amount deposited in bank = Rs.(3 - x) lakhs

According to the question,

${x × 10 × 1}/{100 × 12} + {(3 - x) × 6 × 1}/{100 × 12}$

= $2000/100000 = 1/50$

10x + 18 - 6x

= $1/50$ × 1200 = 24

4x = 24 - 18 = 6

x = $6/4$ = Rs.$3/2$ lakhs

∴ Required difference = 0

Q-9)   The income of a company increases 20% per year. If the income is Rs. 26,64,000 in the year 2012, then its income in the year 2010 was :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Let the income of company in 2010 be Rs.P

According to the question,

A = P$(1 + R/100)^T$

2664000 = P$(1 + 20/100)^2$

2664000 = P$(1 + 1/5)^2$

2664000 = P × $(6/5)^2$

P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000

Q-10)   The income of a company increases 20% per year. If the income is Rs. 26,64,000 in the year 2012, then its income in the year 2010 was :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Let the income of company in 2010 be Rs.P

According to the question,

A = P$(1 + R/100)^T$

2664000 = P$(1 + 20/100)^2$

2664000 = P$(1 + 1/5)^2$

2664000 = P × $(6/5)^2$

P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000